CSCI 4260 MATH 4150

1. CSCI 4260MATH 4150  GRAPH THEORY

2. Last time We learned about BFS We will cover DFS after we learn more about connectivity and digraphs Today, connectivity properties

3. Robustness of connectivity We can make a graph disconnected by deleting either edges or vertices An edge is a bridge of G if G-{e} is disconnected

4. Theorem An edge e is a bridge of G iff e lies on no cycle on G e is a bridge ? e is on no cycle e is on a cycle ? e is not a bridge e is on no cycle ? e is a bridge e is not a bridge ? e is on a cycle

5. bridges In a sense, if a graph contains a bridge, then it�s connectivity is not very robust Can you come up with a graph where every edge is a bridge?

6. Cut vertices v is a cut vertex if G-{v} is disconnected Note that we delete edges incident to v as well

7. Theorem Let v be a vertex incident with a bridge.Then, v is a cut vertex iff deg(v) ? 2 v is a cut vertex ? deg(v) ? 2 deg(v) < 2 ? v is a not cut vertex deg(v) ? 2 ? v is a cut vertex

8. Corollary Given connected G of order ? 3. If G contains a bridge, then G contains a cut vertex

9. Theorem A vertex v of a connected graph G is a cut vertex of G iff there exists two vertices u and w (not equal to v) such that v lies on every u-w path.

10. Non-cut vertices All edges of a tree are bridges Can you come up with a graph where every vertex is a cut vertex?

11. Theorem Let G be a nontrivial, connected graph. Pick a vertex u. If v is a vertex that is farthest from u in G, then v is not a cut vertex.

12. Corollary Every non trivial graph contains at least two non-cut vertices.

13. Non-separable graphs A graph is non-separable if it contains no cut vertices Example?

14. Theorem A graph of order at least 3 is non-separable if and only if every two vertices lie on a common cycle Common cycle ? non-separable Suppose not Non-separable ? common-cycle Suppose not. Pick u and v that do not lie on a common cycle. Among all possible pairs, pick two such that d(u,v) is as small as possible.

15. Non-separable ? common-cycle Case 1: d(u,v) = 1. Not possible. The edge (u,v) is a bridge

17. Blocks Maximal non-separable subgraphs If G is non-separable then it has only one block Analogy: disconnected graphs are composed of components (connected subgraphs)

18. Theorem Define a relation R on edges where e R f : either e=f or  e and f lie on a common cycle. R is an equivalence relation

19. Transitivity

20. Corollary Every two distinct blocks B1 and B2 of a nontrivial connected graph have B1 and B2 are edge disjoint B1 and B2 have at most one vertex in common If v is the common vertex, then v is a cut vertex.

21. B1 and B2 are edge disjoint Immediate because blocks correspond to the equivalence classes of R as we just saw

22. B1 and B2 have at most one vertex in common Suppose not.

23. If v is the common vertex, then v is a cut vertex.

24. Vertex-cuts A set U of vertices is a vertex cut if G-U is disconnected

25. Connectivity of G Minimum vertex cut Define the connectivity of G, ??(G)0 ? ?(G) ? n-1 We say a graph is k-connected if ?(G) ? k

26. Edge-cuts Are defined similarly Minimal vs. minimum?

27. Edge-connectivity?? ?(G): size of a min edge cut G is k-edge connected if ?(G) ? k

28. Theorem For every graph G: ?(G) ? ?(G) ? ?(G)

29. Proof First, special cases G is disconnected or trivial ?(G) = ?(G) = 0 G is complete ?(G) = ?(G) = ?(G) = n -1 We may assume that ?(G) ? n-2

30. Proof (cont) First: ?(G) ? ?(G) Next: ?(G) ? ?(G) Case 1: Let X be a min. edge cut |X| = ?(G) ? n-2 G � X has two components, say G1 and G2 Let k be the order of G1. So order(G2) = n-k Note that every edge in X joins a vertex in G1 to a vertex in G2

31. Order(G1) = k, Order(G2) = n-k, k?1 and n-k?1 Case 1: Every vertex in G1 is adjacent to every vertex in G2 |X| = k(n-k) We have (k-1)(n-k-1) ? 0. Hence, (k-1)(n-k-1) = k(n-k)-n+1 ? 0 Hence ?(G) =k(n-k) ? n-1 But we are studying the case ?(G) ? n-2 So case 1 is not possible

32. Case 2 There exists u in G1 and v in G2 that are not adjacent in G Let us find a vertex cut U of size <= ?(G) For each edge of the edge cut, we add a vertex to U

33. For this last case ?(G) ? |U| ? |X| = ?(G). Done.

34. Interesting cases: For a cubic graph G, ?(G) = ?(G). ?(G) ? ?(G) ?? ?(G) = 3 So there are 4 possible cases. ?(G) = 0: Only if G is disconnected. ?(G) = 0 ?(G) = 3: ?(G) = 3 Other cases: Let U be a min vertex cut; G1 and G2 be the two components in G-U|U| is either 1 or 2. In either case, for every u?U, one of G1 or G2 contains exactly one neighbor of u

35. ?(G) = 1

36. ?(G) = 2

37. Other interesting cases: Last time we proved that, ?(G) ? 2m/n It turns out that this bound is sharp. Define H(r,n): Harary graphs of order n and ? = r

38. A deeper look into connectivity A set S of vertices separate u and v, if G-S is disconnected, and u and v are in different components Note: If S separates u and v S is a vertex cut u and v are not adjacent

39. Paths between u and v Let P = {u, u1, u2, �, uk, v} be a u-v path u1, u2, �, uk are internal vertices A collection P1, P2, .., Pk of u-v paths are internally disjoint if every two paths have no internal vertices common.

40. Menger�s Theorem Let u and v be nonadjacent vertices in a graph G. The minimum number of vertices in a u-v separating set equals the maximum number of internally disjoint u-v paths in G

41. Proof by induction on the size (# of edges) Basis: m = 0. Inductive step. Assume true for all graphs of size ? m Let U be a minimum u-v separating set. Clearly, the number of u-v disjoint paths is at most |U| = k. So we need to show equality.

42. Proof (cont) We look at all minimum u-v separating sets. There are three cases Case 1: There is a u-v separating set U that contains a vertex that is adjacent to both u and v Case 2: There is u-v separating set W with a vertex not adjacent to u and a vertex not adjacent to v What is the remaining case? Case 3: For each min. u-v separating set S, either (every vertex in S is adjacent to u but not to v) or (every vertex in S is adjacent to v but not to u)

43. Case 1 Consider G-{x}: It�s size is less than m U-{x} is a min. separating set for G-{x}. Why? Since |U-{x}| = k -1, by the ind. hypothesis,there are k-1 internally disjoint u-v paths in G-{x} So in G, we have these paths plus u-x-v Done.

44. Case 2 Note: x and y can be the same vertex

45. Case 2 W = {w1, �, wk} First let�s construct G(u) which contains all u-wi paths for all wi ? W in G1 + W (this must be added to the proof in the book) Make a new graph G�(u) by adding a new vertex v� to G(u) and connecting it to all wi Construct G(v) and G�(v) similarly

46. Case 2 size G�(u) < m W is a min u-v� seperating set of size k. By the ind. hyp., there are k disjoint u-v� paths. We take these paths and delete v� from them. Call the resulting paths P1 With similar reasoning, we conclude that G�(v) has k disjoint v-u� paths. Generate paths P2 in a similar fashion. Combine P1 and P2 using the vertices wi. We obtained k internally disjoint paths for G

47. Case 3 We have either the situation on the left or the symmetric case (where v is connected to all in S)

48. Case 3 Let P = {u,x,y, � , v} be a u-v geodesic in G Let e = (x,y) and consider G-e Claim: The size k� of any minimum u-v separating set in G-e is also k. Clearly, k� ? k-1. Why? Suppose, for contradiction, that k� = k-1 (i.e. the claim is false). Let Z be a min u-v separating set in G-e Z + {x} is a min u-v separating set in G So all vertices in Z are adjacent to u (we are in case 3) Z + {y} is a min u-v separating set in G So y is adjacent to v Can you see a contradiction?

49. Case 3 (cont) G-{e} has a min. u-v separating set of size k. By ind. hyp. It has k internally disjoint u-v paths. So does G!

50. We have just proved: Let u and v be nonadjacent vertices in a graph G. The minimum number of vertices in a u-v separating set equals the maximum number of internally disjoint u-v paths in G

51. Theorem A non-trivial graph G = (V,E) is k-connected (k ? 2) if and only if for each distinct u,v ?V there are at least k internally disjoint u-v paths in G

52. k-connected ? k disjoint u-v paths ?u,v We need to prove for any u,v Case 1: u and v are not adjacent Follows from Menger�s theorem Case 2: u and v are adjacent. Let e = (u,v) Claim: G-{e} is (k-1)-connected Then, there are (k-1) internally disjoint paths in G-{e}. In G, we also have the path u-v

53. ?u,v; k disjoint u-v paths ? k-connected Take a min-vertex cut U. Pick u and v from different components There are at least k disjoint u-v paths Menger�s theorem says that the min. size of a u-v separating set is at least k So |U| is at least k The graph is k-connected.

54. Corollary Let G be a k-connected graph and S be any set of k vertices. If a graph H is obtained by adding a new vertex w to G and connecting w to all the vertices in S, then H is also k-connected.

55. Corollary If G is a k connected graph and u, v1, �, vk are k+1 distinct vertics of G, then there exist internally disjoint u-vi paths for all i = 1, �, k

56. Another way of looking at k-connected graphs Last lecture we proved: �A graph of order at least 3 is nonseparable iff every two vertices lie on a common cycle� Note that non-separable means 2 connected. We now prove a generalization of this theorem

57. Theorem If G is a k-connected graph, k ? 2, then every k-vertices of G lie on a common cycle. Proof: suppose not. Suppose there exists S = {v1, �, vk} and there is no cycle containing S. Among all cycles in S, pick cycle C that contains most number of vertices of S.

58. Proof (cont) Suppose C contains l < k vertices from S. Once we choose C, we may assume that the vertices v1, �, vl appear on C in this order (otherwise we rename the vertices).

59. Proof (cont)